1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I When to use which dimensionless number

  1. Jan 3, 2018 #51
    We are looking for the unit vector in the same direction as a super theta. You can’t just divide a vector by its magnitude and use the same symbol for it. Plus, if you are dividing a super theta by its own magnitude to get the unit vector, you need to do the algebra correctly and you need to use a new symbol for it.
     
  2. Jan 3, 2018 #52

    joshmccraney

    User Avatar
    Gold Member

    In post 37 you say:
    and in post 47 you show how to compute the magnitude of ##\hat{a}^{\theta}##. What I tried doing in post 48 was dividing ##\hat{a}^{\theta}## by it's magnitude, making it a unit vector.

    Also, did I use the same symbol? Thought I used your suggestion ##\hat{u}^{\theta}##.
     
  3. Jan 3, 2018 #53

    joshmccraney

    User Avatar
    Gold Member

    Yea, I dropped a term in the algebra. We should have
    $$
    \hat u^\theta
    =\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$
     
  4. Jan 3, 2018 #54
    Right. Now, by extension, please write down what ##\hat{u}^z## would be.
     
  5. Jan 3, 2018 #55

    joshmccraney

    User Avatar
    Gold Member

    Without showing any of the legwork since it's the same as your approach, we have $$\hat{u}^z = \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$
    Do you agree?
     
  6. Jan 4, 2018 #56
    Yes. I'll be back later to continue.
     
  7. Jan 4, 2018 #57
    To refresh your memory, we are going to be performing a force balance over a "window" in the free surface formed by the intersection of planes of constant ##\theta## and z with our free surface. These intersections form grid lines on our free surface. The window will be situated between z and z + dz, and between ##\theta## and ##\theta+d\theta##. In terms of ##g_{zz}##, ##g_{\theta z}##, ##g_{\theta \theta}##, dz, and ##d\theta##, what are the lengths of the four sides of our window?
     
  8. Jan 12, 2018 #58

    joshmccraney

    User Avatar
    Gold Member

    I'm not quite sure what to write. I'm trying to compare what we have to a simpler problem. Geometrically, if we were in cylindrical coordinates the side lengths would be ##r\Delta\theta,\Delta r,\Delta r,(r+\Delta r)\Delta \theta##. This makes me think the lengths should be something like ##a \Delta z, b \Delta z, c \Delta \theta, d \Delta \theta##. The ##g##-subs (as oppose to -supers) were dot products of the in plane vectors ##a##-subs (not necessarily orthonormal). I'm stuck; any hints?

    To make sure I understand what we've done, we started by finding two vectors that lie tangent in the plane and call them ##a##-sub. Then we carefully aligned them with our selected coordinate system and called them ##a##-super. Then we normalized them and called them ##u##. Right?
     
  9. Jan 13, 2018 #59
    We found two vectors in the tangent plane that align with the curved grid lines of constant theta and constant z within the surface. These vectors, when multiplied by the corresponding differentials in the coordinates represent actual differential position vectors within the surface in terms of both their spatial length and their direction. Then we found the a-super vectors normal to the grid lines, with the characteristic that, when dotted with the subs, the dot product is equal to 1 or zero. We then used this to find the equations for unit vectors in the plane of the surface that are normal to the grid lines.

    Regarding our little window in the surface, the sides of the window have differential position vectors along their edges (from corner to corner) of ##\mathbf{a}_{\theta}d\theta## and ##\mathbf{a}_{z}dz##. So, what are the lengths of these edges?
     
  10. Jan 14, 2018 #60

    joshmccraney

    User Avatar
    Gold Member

    Would the sides be ##\sqrt{g_{\theta \theta}}|_zd\theta##, ##\sqrt{g_{\theta \theta}}|_{z+dz}d\theta##, ##\sqrt{g_{zz}}|_\theta dz ##, ##\sqrt{g_{zz}}|_{\theta+d\theta}dz##?
     
  11. Jan 14, 2018 #61
    Josh!!! You're the man!!! Excellent.

    Now we are ready to do a force balance on our little window. In terms of the unit normal u's, what are the in plane tensile forces due to surface tension on the 4 edges of our window (recall that surface tension acts normal to the sides of our window)? (There is also a pressure force acting perpendicular to our window that we'll get to soon)
     
  12. Jan 16, 2018 #62

    joshmccraney

    User Avatar
    Gold Member

    Thanks!

    I think they would be $$\sigma \sqrt{g_{\theta \theta}}|_zd\theta\, \hat u^z\\
    \sigma\sqrt{g_{\theta \theta}}|_{z+dz}d\theta\, \hat u^z\\
    \sigma\sqrt{g_{zz}}|_\theta dz\, \hat u^\theta \\
    \sigma\sqrt{g_{zz}}|_{\theta+d\theta}dz\, \hat u^\theta$$
     
  13. Jan 17, 2018 #63
    Good, except that the ones at evaluated z and theta should be in the negative u directions. Now substitute your equations for the u's into these equations. What do you get?

    Given the equations for the differential position vectors along the edges of the window, can you write a vector equation for the area of the window, including the normal vector to the window?
     
  14. Jan 17, 2018 #64

    joshmccraney

    User Avatar
    Gold Member

    $$\left.-\sigma \sqrt{g_{\theta \theta}}d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z = \left.-\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z\\

    \left.\sigma\sqrt{g_{\theta \theta}}d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz} = \left.\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz}\\

    -\left.\sigma\sqrt{g_{zz}} dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta} = -\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta}\\

    \left.\sigma\sqrt{g_{zz}} dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta} = \left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta}$$

    I think this would be
    $$||\sqrt{g_{\theta \theta}}|_zd\theta\, \hat u^z \times \sqrt{g_{zz}}|_\theta dz\, \hat u^\theta|| \hat u^z \times \hat u^\theta = \sqrt{g_{\theta \theta}}|_zd\theta\cdot \sqrt{g_{zz}}|_\theta dz \, \hat u^z \times \hat u^\theta$$
     
  15. Jan 17, 2018 #65
    It's much simpler than this. The area of the would be ##\hat{a}_{\theta}\times \hat{a}_z d\theta dz##. So the pressure force acting on the window from inside to outside would be ##p\hat{a}_{\theta}\times \hat{a}_z d\theta dz##, where p is the gauge pressure inside the fluid at the interface.
     
  16. Jan 18, 2018 #66

    joshmccraney

    User Avatar
    Gold Member

    Of course, makes sense! I'm with you. So we've accounted for surface tension and pressure.
     
  17. Jan 19, 2018 #67
    Are you saying there are other forces acting on our window?
     
  18. Jan 19, 2018 #68

    joshmccraney

    User Avatar
    Gold Member

    Well, we're ignoring gravity. Viscous forces though from the side walls, what do you think?
     
  19. Jan 19, 2018 #69
    I think that the interface has no mass, so that the gravitational force on the window is zero. The viscous contribution to the normal stress at the interface, when lumped in with the pressure (the shear stress is, of course, zero) might contribute a little, and we should consider that later (although I don't think it would be important). It should be neglected for now.
     
  20. Jan 19, 2018 #70

    joshmccraney

    User Avatar
    Gold Member

    I'm with you, and agree with everything you've said so far!
     
  21. Jan 20, 2018 #71
    OK. Please write down the force balance on the window, and divide by dtheta and dz. What do you get?
     
  22. Jan 20, 2018 #72

    joshmccraney

    User Avatar
    Gold Member

    $$
    \left.\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz} \left.-\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z+
    \left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta}-\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta}+
    p\hat{a}_{\theta}\times \hat{a}_z d\theta dz = \vec 0\\

    \frac{\partial}{\partial z}\left(\sigma \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
    \frac{\partial}{\partial \theta}\left(\sigma \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
    p\hat{a}_{\theta}\times \hat{a}_z = \vec 0
    $$

    where the rhs is 0 since the membrane does not have any mass. Something doesn't seem right though: the pressure term is the only term that is normal to the surface. If my work above is correct, then there would not be anything to balance it, implying ##p=0##. What do you think?
     
  23. Jan 20, 2018 #73
    You can factor out the ##\sigma## from the derivatives. Regarding your question, the spatial derivatives of the coordinate basis vectors ##\hat{a}_{\theta}## and ##\hat{a}_z## along the surface have components normal to the surface. This is what the pressure has to balance.

    In anticipation of what we do next, what is the magnitude of the vector ##\hat{a}_{\theta}\times \hat{a}_z##?
     
  24. Jan 29, 2018 #74

    joshmccraney

    User Avatar
    Gold Member

    From your previous post we have $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}\implies \\
    \hat{a}_\theta \times \hat{a}_z = \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right) \times \left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\
    =-R \frac{\partial R}{\partial z}\hat z-\frac{\partial R}{\partial \theta}\hat \theta+R\hat r \implies\\
    ||\hat{a}_\theta \times \hat{a}_z|| = \sqrt{\left( R \frac{\partial R}{\partial z} \right)^2+\left( \frac{\partial R}{\partial \theta} \right)^2+R^2}$$
    Geometrically this would be the area of a window along the surface. Is this what you're looking for?
     
  25. Jan 29, 2018 #75
    It's also equal to ##\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}##. Can you show this?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted